Physics quiz question: "10 bar" depth underwater

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA

"JLC Tribute to Deep Sea Alarm, caseback"
Peter Chong, Deployant
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A diving watch claims to have a waterproof rating of 
"10 bar," meaning that it can be safely submerged to withstand an increase of 1.0×10⁶ Pa of pressure compared to that at the surface.[*]

Starting from the surface, this watch would need to be taken __________ downwards into fresh water to experience a pressure increase of 1.0×10⁶ Pa. (ρ_water = 1.0×10³ kg/m³.)
(A) 10 m.
(B) 1.0×10² m.
(C) 7.8×10⁴ m.
(D) 1.0×10⁵ m.

[*] deployant.com/comparative-review-jlc-memovox-deep-sea-vulcain-nautical-seventies-part-1-2/.

Correct answer (highlight to unhide): (B)

For static fluids, the energy density relation between pressure difference ∆P and difference in depth ∆y is given by:

0 = ∆P + ρ_water·g·∆y.

Since the difference in pressure ∆P = 1.0×10⁶ Pa, then the depth in water beneath the surface can be solved for:

–ρ_water·g·∆y = ∆P,

∆y = –∆P/(ρ_water·g) = –(1.0×10⁶ Pa)/((1.0×10³ kg/m³)·(9.80 m/s²)),

∆y = –102.0408163265... m,

where the negative sign is understood as the depth below the surface of the water, and to two significant figures, this depth is 1.0×10² m.

(Response (A) is P_atm/(ρ·g); response (C) is –∆P/(ρ_air·g); while response (D) –∆P/g.)

Sections 70854, 70855
Exam code: quiz05Ro74
(A) : 7 students
(B) : 34 students
(C) : 5 students
(D) : 6 students

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63